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Chapter 5: Problem 105

Hydrogen azide, \(\mathrm{HN}_{3},\) decomposes on heating by the following unbalanced equation: $$\mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g)$$ If 3.0 atm of pure \(\mathrm{HN}_{3}(g)\) is decomposed initially, what is the final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.

Short Answer

Expert verified
The final total pressure in the reaction container is 6.0 atm, with partial pressures of nitrogen gas and hydrogen gas being 1.5 atm and 4.5 atm, respectively.

Step by step solution

01

Balance the chemical equation

First, we need to balance the given chemical equation: $$\mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g)$$ Balancing the equation gives: $$2\mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g)$$
02

Determine mole ratios

From the balanced equation, we can determine the mole ratios of the reactants and products: $$ \begin{cases} 2\, moles\, of\, HN_{3} \longrightarrow 1\, mole\, of\, N_{2} \\ 2\, moles\, of\, HN_{3} \longrightarrow 3\, moles\, of\, H_{2} \end{cases} $$
03

Determine change in moles

Since the initial pressure of HN3 is 3.0 atm and the volume and temperature are constant, we can use the mole ratios from step 2 to determine the change in moles for each substance: Change in moles of HN3: -3 atm (decreases as it decomposes) Change in moles of N2: +1.5 atm (half of the HN3 decomposes into N2) Change in moles of H2: +4.5 atm (1.5 times the HN3 decomposes into H2)
04

Calculate final total pressure

Now we add the changes in moles to the initial pressure to find the final total pressure: Final total pressure = Initial HN3 pressure + Change in N2 pressure + Change in H2 pressure Final total pressure = 3.0 atm - 3.0 atm + 1.5 atm + 4.5 atm = 6.0 atm
05

Calculate partial pressures

Finally, we can calculate the partial pressures of nitrogen and hydrogen gas using the mole fractions and total pressure: Partial pressure of N2 = (N2 moles / Total moles) * Total pressure Partial pressure of N2 = (1.5 / 6.0) * 6.0 atm = 1.5 atm Partial pressure of H2 = (H2 moles / Total moles) * Total pressure Partial pressure of H2 = (4.5 / 6.0) * 6.0 atm = 4.5 atm The final total pressure in the reaction container is 6.0 atm, the partial pressure of nitrogen gas is 1.5 atm, and the partial pressure of hydrogen gas is 4.5 atm.

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Most popular questions from this chapter

In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density $=1.0 \mathrm{g} / \mathrm{cm}^{3} )$ is allowed to take place in a bottle with a total volume of 825 \(\mathrm{mL}\) until 12\(\%\) by volume is ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( . Assuming that the \)\mathrm{CO}_{2}$ is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is $0.79 \mathrm{g} / \mathrm{cm}^{3} . )$

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\) . The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{K} .\) What is the final volume of the balloon?

Consider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}( )$$ Suppose a 32.9 -g sample of \(\mathrm{CaSiO}_{3}\) is reacted with 31.8 \(\mathrm{L}\) of \(\mathrm{HF}\) at \(27.0^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) . Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

The average lung capacity of a human is 6.0 L. How many moles of air are in your lungs when you are in the following situations? a. At sea level \((T=298 \mathrm{K}, P=1.00 \mathrm{atm})\) b. \(10 . \mathrm{m}\) below water \((T=298 \mathrm{K}, P=1.97 \mathrm{atm})\) c. At the top of Mount Everest \((T=200 . \mathrm{K}, P=0.296 \mathrm{atm})\)

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 \(\mathrm{K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 \(\mathrm{K}\). This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon
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